Problem: Evaluate the triple integral. $ \int_1^2 \int_1^z \int_{-y}^y 2x - z \, dx \, dy \, dz =$
Solution: We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_1^2 \int_1^z \int_{-y}^y 2x - z \, dx \, dy \, dz \\ \\ &= \int_1^2 \int_1^z \left[ x^2 - zx \right]_{-y}^y dy \, dz \\ \\ &= \int_1^2 \int_1^z (y^2 - zy) - (y^2 + zy) \, dy \, dz \\ \\ &= \int_1^2 \int_1^z -2zy \, dy \, dz \end{aligned}$ The second layer: $\begin{aligned} &\int_1^2 \int_1^z -2zy \, dy \, dz \\ \\ &= \int_1^2 \left[ -zy^2 \right]_1^z dz \\ \\ &= \int_1^2 -z^3 + z \, dz \end{aligned}$ The third layer: $\begin{aligned} &\int_1^2 -z^3 + z \, dz \\ \\ &= \left[ \dfrac{-z^4}{4} + \dfrac{z^2}{2} \right]_1^2 \\ \\ &= (-4 + 2) - \left( \dfrac{-1}{4} + \dfrac{1}{2} \right) \\ \\ &= -2 - \dfrac{1}{4} \\ \\ &= \dfrac{-9}{4} \end{aligned}$ In conclusion: $ \int_1^2 \int_1^z \int_{-y}^y 2x - z \, dx \, dy \, dz = \dfrac{-9}{4}$